Linear Momentum and Collisions

55 Elastic Collisions in One Dimension

Learning Objectives

  • Describe an elastic collision between two objects in one dimension.
  • Define internal kinetic energy and its conservation.
  • Derive the condition for conservation of internal kinetic energy.
  • Determine final velocities of two objects in an elastic collision given masses and initial velocities.
Collisions occur in many biological and physical systems, from molecules colliding in a gas to cells interacting in a bloodstream. Understanding how momentum and energy are transferred in these events helps us understand many real-world processes. In this section, we begin with the simplest type of collision: a one-dimensional interaction between two objects. We specifically focus on elastic collisions, which are idealized events in which both momentum and internal kinetic energy are conserved. Internal kinetic energy is the total kinetic energy of all objects within a system. In an elastic collision, this total remains unchanged before and after the event.

Elastic Collision

An elastic collision is one in which both momentum and internal kinetic energy are conserved.

Internal Kinetic Energy

Internal kinetic energy refers to the total kinetic energy of all objects in a system. It does not include potential energy or energy lost to sound, heat, or deformation. Elastic collisions are common in the microscopic world. For example, electrons colliding with atomic nuclei can be very nearly perfectly elastic. At the macroscopic scale, however, some kinetic energy is usually converted into sound, thermal energy, or deformation. But under carefully controlled conditions, such as when two steel blocks slide on an icy surface or when two carts with spring bumpers collide on a nearly frictionless air track, we can closely approximate elastic behavior. Figure 55.1 shows a classic one-dimensional elastic collision: two objects interact on a frictionless surface, conserving both momentum and internal kinetic energy.

The system of interest contains a smaller mass m sub1 and a larger mass m sub2 moving on a frictionless surface. M sub 2 moves with velocity V sub 2 and momentum p sub 2 and m sub 1 moves behind m sub 2, with velocity V sub 1 and momentum p sub 1 toward the right direction. P 1 plus P 2 equals p total. The net force is zero. After collision m sub 1 moves toward the left with velocity V sub 1 while m sub 2 moves toward the right with velocity V sub 2 on the same frictionless surface. The momentum of m sub 1 becomes p 1 prime and m 2 becomes p 2 prime now. P 1 prime plus p 2 prime equals p total.
Figure 55.1: Internal kinetic energy is the sum of the kinetic energies of the objects in the system. An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved.
To analyze elastic collisions, we use two conservation laws: conservation of momentum and conservation of internal kinetic energy. The conservation of momentum for two objects in one dimension is written as:
[latex]{p}_{1} + {p}_{2} = {p\prime}_{1} + {p\prime}_{2} \quad \text{(if } \mathbf{F}_{\text{net}} = 0\text{)}[/latex]

Expanding this using mass and velocity:

[latex]{m}_{1}{v}_{1} + {m}_{2}{v}_{2} = {m}_{1}{v\prime}_{1} + {m}_{2}{v\prime}_{2}[/latex]

Here, [latex]v_1[/latex] and [latex]v_2[/latex] are the initial velocities of objects 1 and 2, and [latex]v'_1[/latex] and [latex]v'_2[/latex] are their final velocities after the collision. Primes (') denote quantities after the interaction. For an elastic collision, we also conserve internal kinetic energy. This gives us the second equation:

[latex]\frac{1}{2}{m}_{1}v_{1}^{2} + \frac{1}{2}{m}_{2}v_{2}^{2} = \frac{1}{2}{m}_{1}v_{1}^{\prime 2} + \frac{1}{2}{m}_{2}v_{2}^{\prime 2}[/latex]

Together, these two equations allow us to solve for the unknown final velocities in an elastic collision.

Example 55.1: Calculating Velocities Following an Elastic Collision

Calculate the velocities of two objects following an elastic collision, given that

[latex]{m}_{1}=0\text{.}\text{500 kg,}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}{m}_{2}=3\text{.}\text{50 kg,}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}{v}_{1}=4\text{.}\text{00 m/s, and}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}{v}_{2}=0[/latex]

Strategy and Concept

First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 55.1 where both objects are initially moving. We are asked to find two unknowns (the final velocities [latex]{v\prime }_{1}[/latex] and [latex]{v\prime }_{2}[/latex]). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus [latex]{v}_{2}=0[/latex]. Once we simplify these equations, we combine them algebraically to solve for the unknowns.

Solution

For this problem, note that [latex]{v}_{2}=0[/latex] and use conservation of momentum. Thus,

[latex]{p}_{1}=p\prime {}_{1}+p\prime {}_{2}[/latex]

or

[latex]{m}_{1}{v}_{1}={m}_{1}{v\prime }_{1}+{m}_{2}{v\prime }_{2}[/latex]

Using conservation of internal kinetic energy and that [latex]{v}_{2}=0[/latex],

[latex]\frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{1}{v\prime }_{1}{}^{2}+\frac{1}{2}{m}_{2}{v\prime }_{2}{}^{2}[/latex]

Solving the first equation (momentum equation) for [latex]{v\prime }_{2}[/latex], we obtain

[latex]{v\prime }_{2}=\frac{{m}_{1}}{{m}_{2}}\left({v}_{1}-{v\prime }_{1}\right)[/latex]

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable [latex]{v\prime }_{2}[/latex], leaving only [latex]{v\prime }_{1}[/latex] as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are

[latex]{v\prime }_{1}=4\text{.}\text{00 m/s}[/latex]

and

[latex]{v\prime }_{1}=-3\text{.}\text{00 m/s}[/latex]

As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution [latex]\left({v\prime }_{1}=-3\text{.}\text{00 m/s}\right)[/latex] is negative, meaning that the first object bounces backward. When this negative value of [latex]{v\prime }_{1}[/latex] is used to find the velocity of the second object after the collision, we get

[latex]{v\prime }_{2}=\frac{{m}_{1}}{{m}_{2}}\left({v}_{1}-{v\prime }_{1}\right)=\frac{0\text{.}\text{500 kg}}{3\text{.}\text{50 kg}}\left[4\text{.}\text{00}-\left(-3\text{.}\text{00}\right)\right]\phantom{\rule{0.25em}{0ex}}\text{m/s}[/latex]

or

[latex]{v\prime }_{2}=1\text{.}\text{00 m/s}[/latex]

Discussion

The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged.

The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed.

Take-Home Experiment: Ice Cubes and Elastic Collisions

Find two or more ice cubes of similar size and use a smooth surface such as a glass tabletop or kitchen counter. Set the ice cubes several centimeters apart. Gently flick one ice cube so it slides across the surface and collides with another stationary ice cube. Observe what happens after the collision. To get the clearest results, try to avoid glancing or spinning collisions. Instead, aim for direct impacts with little to no rotation. Watch carefully how the ice cubes move before and after the collision. Does the moving ice cube slow down while the stationary one begins to move? Are the total speeds before and after the collision roughly the same? Use what you've learned about momentum conservation to explain the outcome. Since friction is minimal on the cold, smooth surface, these ice cube collisions may approximate elastic collisions. In such cases, both total momentum and internal kinetic energy are nearly conserved.

Interactive Exploration: Collision Lab

Collisions provide an excellent opportunity to study two of the most important conservation laws in physics: the conservation of momentum and, in some cases, the conservation of kinetic energy. In this simulation, you'll investigate collisions between discs on a nearly frictionless surface while changing their masses, speeds, directions, and the elasticity of the collisions. Experiment with different collision scenarios and observe the motion before and after impact. Use the built-in graphs and data displays to compare the total momentum and kinetic energy of the system. As you explore, determine which physical quantities are always conserved and which depend on the type of collision.

Guided Exploration

As you interact with the simulation, try to answer the following questions:

  1. Begin with two discs of equal mass undergoing a head-on collision. Compare the total momentum before and after the collision. What do you observe?
  2. Repeat the experiment using discs with different masses. How do the masses affect the velocities of the discs after the collision?
  3. Change the initial speeds and directions to create an angled, two-dimensional collision. Is the total momentum still conserved?
  4. Adjust the elasticity from perfectly elastic to perfectly inelastic. How does the total kinetic energy change as the elasticity decreases?
  5. Observe the momentum and kinetic energy graphs throughout several collisions. Which quantity is always conserved? Under what conditions is kinetic energy conserved?
  6. Add a third disc to the simulation. How does increasing the number of objects affect the complexity of the motion while still satisfying the conservation laws?

After completing the exploration, compare your observations with the concepts presented in this section. Notice that the total momentum of an isolated system remains constant in every collision, while kinetic energy is conserved only in elastic collisions. These conservation laws are fundamental to understanding interactions ranging from atomic particles to vehicle collisions and sporting events.

Example 55.2: Elastic Collisions Between Gas Molecules in the Lungs

Gas exchange in the alveoli depends on countless molecular collisions, most of which can be treated as elastic: the total kinetic energy of the colliding molecules is conserved, even though individual molecules speed up or slow down. Consider an oxygen molecule (mass 32 u, where 1 u is one atomic mass unit) moving at a typical thermal speed of 490 m/s at body temperature, which collides head-on with a nitrogen molecule (mass 28 u) that is momentarily at rest. Find the velocity of each molecule immediately after the elastic collision.

Strategy

Because the target molecule starts at rest, we can use the standard results for a one-dimensional elastic collision derived earlier in this section:

[latex]{v}_{1}^{\prime }=\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}{v}_{1}\phantom{\rule{0.75em}{0ex}}{v}_{2}^{\prime }=\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}{v}_{1}[/latex]

where molecule 1 is O2 and molecule 2 is N2. Because both masses appear only as a ratio, they can be left in atomic mass units without converting to kilograms.

Solution

The oxygen molecule’s velocity after the collision is:

[latex]{v}_{1}^{\prime }=\frac{32\ \text{u}-28\ \text{u}}{32\ \text{u}+28\ \text{u}}\left(490\ \text{m/s}\right)=33\ \text{m/s}[/latex]

The nitrogen molecule’s velocity after the collision is:

[latex]{v}_{2}^{\prime }=\frac{2\left(32\ \text{u}\right)}{32\ \text{u}+28\ \text{u}}\left(490\ \text{m/s}\right)=520\ \text{m/s}[/latex]

Discussion

Because the two molecular masses are similar, most of the oxygen molecule’s kinetic energy is transferred to the nitrogen molecule, which ends up moving faster than the oxygen molecule’s original speed. Multiply this single collision by the roughly 1023 molecular collisions occurring every second in a single alveolus, and you get the constant, random molecular motion that drives diffusion—the process by which oxygen ultimately moves from inhaled air into the blood, discussed further in the chapter on molecular transport phenomena.

Section Summary

  • An elastic collision is one in which internal kinetic energy is conserved in addition to momentum.
  • When both momentum and kinetic energy are conserved, we can use these principles to calculate the final velocities of the objects involved in a one-dimensional, two-body collision.
  • This analysis is especially useful when studying systems with minimal energy loss, such as collisions between low-friction carts, subatomic particles, or ice cubes on a smooth surface.
  • Although most macroscopic collisions are not perfectly elastic, they can often be approximated as such in controlled environments, allowing for meaningful application of conservation laws.

Conceptual Questions

  1. What is an elastic collision?

Problems & Exercises

  1. Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.
  2. Professional Application: Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of [latex]4\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex], and the second a mass of [latex]7\text{.}\text{50}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex]. If the two satellites collide elastically rather than dock, what is their final relative velocity?
  3. A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?

Glossary

elastic collision
a collision that also conserves internal kinetic energy
internal kinetic energy
the sum of the kinetic energies of the objects in a system
definition

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Introductory Physics for the Health and Life Sciences I Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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